\newproblem{lay:2_2_13}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 2.2.13}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
  Suppose $AB=AC$, where $B$ and $C$ are $n\times p$ matrices and $A$ is invertible. Show that $B=C$. Is this true, in general,
	if $A$ is not invertible?
}{
  % Solution
	If $A$ is invertible we multiply on the left by $A^{-1}$ to obtain
	\begin{center}
		$A^{-1}(AB)=A^{-1}(AC)$ \\
		$(A^{-1}A)B=(A^{-1}A)C$ \\
		$I_nB=I_nC$ \\
		$B=C$
	\end{center}
	If $A$ is not invertible, then the statement is not generally true. For example, let $A=\begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}$, $B=\begin{pmatrix}1\\0\end{pmatrix}$
	and $C=\begin{pmatrix}1\\1\end{pmatrix}$.
	\begin{center}
		$AB=\begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}1\\0\end{pmatrix}=
		    \begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}\begin{pmatrix}1\\1\end{pmatrix}=AC$\\
	\end{center}
}
\useproblem{lay:2_2_13}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
